જો theta = tan^{-1}left(frac{1}{3}right) + ta | vedclass

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(C) આપણે જાણીએ છીએ કે $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$.
દરેક પદને $	an^{-1}\left(\frac{1}{n^2+n+1}ight) =

Vedclass · Accepted Answer

$\frac{5}{7}$

Vedclass · Answer

(C) આપણે જાણીએ છીએ કે $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$.
દરેક પદને $	an^{-1}\left(\frac{1}{n^2+n+1}ight) = 	an^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}ight) = 	an^{-1}(n+1) - 	an^{-1}(n)$ તરીકે દર્શાવી શકાય છે.
આપેલા પદો પર આ લાગુ પાડતા:
$	an^{-1}\left(\frac{1}{3}ight) = 	an^{-1}(2) - 	an^{-1}(1)$
$	an^{-1}\left(\frac{1}{7}ight) = 	an^{-1}(3) - 	an^{-1}(2)$
$	an^{-1}\left(\frac{1}{13}ight) = 	an^{-1}(4) - 	an^{-1}(3)$
$	an^{-1}\left(\frac{1}{21}ight) = 	an^{-1}(5) - 	an^{-1}(4)$
$	an^{-1}\left(\frac{1}{31}ight) = 	an^{-1}(6) - 	an^{-1}(5)$
આનો સરવાળો કરતા,આપણને ટેલિસ્કોપિંગ શ્રેણી મળે છે:
$	heta = (	an^{-1}(2) - 	an^{-1}(1)) + (	an^{-1}(3) - 	an^{-1}(2)) + (	an^{-1}(4) - 	an^{-1}(3)) + (	an^{-1}(5) - 	an^{-1}(4)) + (	an^{-1}(6) - 	an^{-1}(5))$
$	heta = 	an^{-1}(6) - 	an^{-1}(1)$
સૂત્ર $	an^{-1}(x) - 	an^{-1}(y) = 	an^{-1}\left(\frac{x-y}{1+xy}ight)$ નો ઉપયોગ કરતા:
$	heta = 	an^{-1}\left(\frac{6-1}{1+6 	imes 1}ight) = 	an^{-1}\left(\frac{5}{7}ight)$
તેથી,$	an 	heta = \frac{5}{7}$.

જો $\theta = \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{13}\right) + \tan^{-1}\left(\frac{1}{21}\right) + \tan^{-1}\left(\frac{1}{31}\right)$,તો $\tan \theta =$

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